# help me echelon matrix ?

9 views (last 30 days)
Nguyen Trong Nhan on 5 Jan 2014
Answered: Roger Stafford on 6 Jan 2014
I have a matrix:
syms m
A = [1 2 3 4;2 -1 1 1;-1 4 3 2;1 2 0 m]
with m is the parameter I use rref command:
rref(A)
ans =
[ 1, 0, 0, 0]
[ 0, 1, 0, 0]
[ 0, 0, 1, 0]
[ 0, 0, 0, 1]
the parameter m was lost. How I can convert matrix A to echelon form that the parameter m is still kept. thanks you very much.

Matt J on 5 Jan 2014
Edited: Matt J on 5 Jan 2014
I suspect, after experimenting with multiple m, that the echelon form of this particular matrix is independent of m, e.g.,
>> m=1; A = [1 2 3 4;2 -1 1 1;-1 4 3 2;1 2 0 m]; rref(A)
ans =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
>> m=10; A = [1 2 3 4;2 -1 1 1;-1 4 3 2;1 2 0 m]; rref(A)
ans =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

Roger Stafford on 6 Jan 2014
It is in the nature of the reduced row echelon form for your square matrix to be the identity matrix. That is because no matter what value m has, the matrix is always non-singular.
Suppose we alter one number in A:
A = [1 2 3 4;2 -1 1 1;1 4 3 2;1 2 0 m]
Then rref(A) still gives the appearance of being independent of the value of m. However there is one and only one value, m = -3.2, which makes the matrix singular and in that case the bottom row of rref(A) will become all zeros, showing that it can be affected by the value of m.
http://en.wikipedia.org/wiki/Row_echelon_form

### Categories

Find more on Linear Algebra in Help Center and File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by